Segmention fault with sizeof as dimension of string

Question

I'm writing a little program to order a matrix of strings. In the ordination algorithm I use this function to swap two strings, but in same cases the program crashhes with Segmentation Fault Error. I have understood it's caused by how I inizialize the temp string, but I haven't understood why.

void stringSwap(char*string1,char*string2)
{   const int dim=sizeof(string1);
    char temp[dim];
    strcpy(temp,string1);
    strcpy(string1,string2);
    strcpy(string2,temp);
}

Can someone explain me why it gives this error? And is there another mode to do this thing correctly without using the dinamic allocation that I don't know actually? Thank you very much!

You should be using [strlen](http://en.cppreference.com/w/cpp/string/byte/strlen) not `sizeof`. — Shafik Yaghmour, Feb 12 '14 at 20:22
sizeof does not work the way you want. Also standard C++ does not have variable length arrays. — drescherjm, Feb 12 '14 at 20:22

score 3 · Answer 1 · answered Feb 12 '14 at 20:22

string1 is a pointer (not an array, even if arrays are decayed into pointers), and the sizeof some pointer is fixed (it is usually the word size in bytes, e.g. 4 bytes on 32 bits machines, and 8 bytes on 64 bits machines & systems).

The sizeof some pointer is a compile-time constant and does not depend upon the value of that pointer.

score 3 · Accepted Answer · answered Feb 12 '14 at 20:22

3

sizeof(string1) is 4 or 8 bytes, depending on your system.

You should use strlen(string1)+1 instead.

answered Feb 12 '14 at 20:22

barak manos

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score 1 · Answer 3 · answered Feb 12 '14 at 20:27

This statement

const int dim=sizeof(string1);

will return sizeof of pointer to char that is sizeof( char * ). It is usually equal to either 4 or 8 bytes depending on the platform where you run your program.

I think either the both strings shall have the same size (they can have different actual lengths but sizes shall be equal) and you have to pass that size to the function or this swap function has no sense.

ak_2050 · Answer 4 · 2014-02-12T20:32:13.100

0

try:

void stringSwap(char*string1,char*string2)
{   
    char temp[strlen(string1)+1];
    strcpy(temp,string1);
    strcpy(string1,string2);
    strcpy(string2,temp);
}

edited Feb 12 '14 at 20:32

answered Feb 12 '14 at 20:26

ak_2050

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4

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Your code has undefined behaviour. You may not pass o to standard C string function. – Vlad from Moscow Feb 12 '14 at 20:28
thanks man. i edited. char *temp = "" is what I meant originally. – ak_2050 Feb 12 '14 at 20:34
The size of an array shall be a constant expression in C++. So this statement char temp[strlen(string1)+1]; is invalid. – Vlad from Moscow Feb 12 '14 at 20:38
Oh, I am sorry. The qiestion is marked as C question not C++. – Vlad from Moscow Feb 12 '14 at 20:40
no worries. i know what you meant. i'm a big C++ guy as well. If this was in C++ this whole functions could just be: string1.swap(string2); – ak_2050 Feb 12 '14 at 20:45
The question was originally tagged as c++. – drescherjm Feb 12 '14 at 20:48

Segmention fault with sizeof as dimension of string

4 Answers4